How do you identify all asymptotes or holes for #f(x)=(x^2-9)/(2x^2+1)#?

1 Answer
Sep 17, 2016

Answer:

horizontal asymptote at #y=1/2#

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve: #2x^2+1=0rArrx^2=-1/2#

there are no real solutions for x hence there are no vertical asymptotes.

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" (a constant)"#

divide terms on numerator/denominator by the highest power of x that is #x^2#

#f(x)=(x^2/x^2-9/x^2)/((2x^2)/x^2+1/x^2)=(1-9/x^2)/(2+1/x^2)#

as #xto+-oo,f(x)to(1-0)/(2+0)#

#rArry=1/2" is the asymptote"#

Holes occur when there are duplicate factors on the numerator/denominator. This is not the case here hence there are no holes.
graph{(x^2-9)/(2x^2+1) [-18.01, 18.04, -9.02, 9]}